Question

我有两个实体：

1）

@Entity
@Table(name = "users")
public class User {

    private String username;
    private String password;
    private String name;
    private String surname;
    @ManyToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "depId")
    private Department department;
    private boolean enabled;
    private Set<UserRole> userRole = new HashSet<UserRole>(0);
    private Set<Equipment> equipment = new HashSet<Equipment>();

2）

@Entity
@Table(name = "department")
public class Department {

    private int depId;
    private String name;
    private String adress;
    @OneToMany(fetch = FetchType.EAGER, mappedBy = "department")
    private Set<User> user = new HashSet<User>();

如何编写HQL查询，将值INSERT到User。不是从其他表。我有JSP表单，其中插入表单。我想从thats表单中添加值。我有'删除'方法的问题。

在UserDAOImpl：

@Override
    public User findByUserName(String username) {
        return (User) getSession().get(User.class, username);
    }

@Override
    public void delete(String username) {
        getSession().delete(findByUserName(username));
    }

当我使用此方法时，它会删除Department中的“department”和UserRole表中的“user_role”。但我想只删除用户。

更新

此问题已修复。但我有第二个问题。我有第二个实体，如Depatment，名为UserRole。并且有相同的方法。我需要方法：

UserRoleDAOImpl：

@Override
public Set<UserRole> findByRole (Set<UserRole> userRole){

}

DepartmentDAOImpl：

@Override
public Department findByDepartmentName(String departmentName) {
    List<Department> departments = getSession()
            .createQuery("select d from Department d where d.name =:departmentName")
            .setParameter("departmentName", departmentName).list();
    return departments.size() == 0 ? null : departments.get(0);
}

控制器：

@RequestMapping(value = "/updateUser", method = RequestMethod.POST)
private ModelAndView updateUser(@ModelAttribute("users") User user, BindingResult result) {
    ModelAndView model = new ModelAndView();
    department = departmentService.findByDepartmentName(user.getDepartment().getName());
    user.setDepartment(department);
    userRole = userRoleService.findByRole(user.getUserRole());
    user.setUserRole(userRole);
    userService.update(user);
    model.addObject("allUsers", userService.getAllUsers());
    model.setViewName("superAdmin");
    return model;
}

我不知道怎么写userRoleService.findByRole(user.getUserRole())。它不能正常工作。

Answer 1

由于您正在使用注释，因此需要阅读有关插入db的hibernate事务。希望您已经在db中创建了一个表。

您可以为应用程序创建SessionFactory对象，并使用Criteria api将表单值保存到db。请查找示例链接

http://www.dineshonjava.com/2013/08/spring-crud-example-using-many-to-one.html#.V8aQp_l97IU

如果您仍然只想编写查询，则可以使用带有hql的条件

https://www.mkyong.com/hibernate/hibernate-criteria-examples/

由于普拉迪普

Hibernate从表单中插入外键

1 个答案: